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Find the arithmetic sequence whose 20th term is 28, given that the sum of its third and sixth terms is greater than its ninth term by eight.
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The first term of any arithmetic sequence is denoted by the letter π.
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And the common difference is equal to π.
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This is the difference between each of the terms.
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The second term of any arithmetic sequence is therefore equal to π plus π.
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Adding another π to this means that the third term is π plus two π.
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This pattern continues such that the πth term is equal to π plus π minus one multiplied by π.
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Weβre told in this question that the 20th term is equal to 28.
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This means that π plus 19π equals 28.
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Weβre also told that the sum of the third and sixth terms is greater than the ninth term by eight.
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This means that the third term plus the sixth term is equal to the ninth term plus eight.
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This can be written as π plus two π plus π plus five π is equal to π plus eight π plus eight.
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Collecting like terms on the left-hand side gives us two π plus seven π.
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Subtracting π plus eight π from both sides of this equation gives us π minus π is equal to eight, as two π minus π is π and seven π minus eight π is negative π.
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We now have a pair of simultaneous equations that we can solve to calculate the values of π and π.
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When we subtract equation two from equation one, the πβs cancel.
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Subtracting negative π is the same as adding π.
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And 19π plus π is equal to 20π.
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28 minus eight is equal to 20.
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Dividing both sides of this equation by 20 gives us a value of π equal to one.
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The common difference of the sequence is one.
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We can then substitute this value into equation one or equation two to calculate the value of π.
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Substituting into equation two gives us π minus one is equal to eight.
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Adding one to both sides of this equation gives us π is equal to nine.
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As the first term is equal to nine and the common difference is one, our arithmetic sequence is nine, 10, 11, and so on.